Chemistry · Aldehydes, Ketones and Carboxylic Acids

Nucleophilic Addition Reactions of Aldehydes & Ketones

The polar carbonyl group, $\ce{>C=O}$, is the reactive heart of every aldehyde and ketone, and almost everything NCERT Unit 8 (§8.4) lists as a "chemical reaction" begins with a nucleophile attacking its electron-poor carbon. This deep dive builds the mechanism from first principles, fixes the reactivity order that NEET tests every year, and then walks through each named addition — cyanohydrin, bisulphite, Grignard, acetal, hydrate, and the addition-elimination products with ammonia derivatives.

Why the carbonyl invites nucleophiles

The carbonyl carbon is $sp^2$-hybridised: it forms three $\sigma$-bonds lying in one plane at roughly $120^\circ$, while its remaining $p$-orbital overlaps side-on with a $p$-orbital of oxygen to give the $\pi$-bond. Oxygen is far more electronegative than carbon, so this $\pi$-cloud is pulled towards oxygen. NCERT describes the bond through resonance between a neutral and a dipolar form, leaving the carbon electron-deficient.

$$\ce{>C=O <-> >\overset{+}{C}-\overset{-}{O}}$$

The consequence is a permanent polarity: the carbonyl carbon carries a partial positive charge ($\delta^+$) and is electrophilic (a Lewis acid), while oxygen carries $\delta^-$ and, with its two lone pairs, is nucleophilic (a Lewis base). Compare this with an alkene, whose non-polar, electron-rich $\ce{C=C}$ attracts electrophiles. The reversed polarity is exactly why aldehydes and ketones undergo nucleophilic addition rather than the electrophilic addition of alkenes.

Three reactive centres exist in a carbonyl molecule: the nucleophilic oxygen (attacked by electrophiles), the electrophilic carbon (attacked by nucleophiles), and the acidic $\alpha$-hydrogen. This page covers the carbon-centred chemistry; the $\alpha$-hydrogen chemistry is treated separately.

The general mechanism

A nucleophile $\ce{Nu^-}$ approaches the planar carbonyl carbon from a direction roughly perpendicular to the plane of the $sp^2$ orbitals. As the new $\ce{C-Nu}$ bond forms, the carbon rehybridises from $sp^2$ to $sp^3$, the $\pi$-electrons shift onto oxygen, and a tetrahedral alkoxide intermediate is produced. This alkoxide then captures a proton from the medium (water or alcohol) to give the neutral addition product. The net change is the addition of $\ce{Nu^-}$ and $\ce{H+}$ across the $\ce{C=O}$ double bond.

Figure 1

Nucleophilic addition to a carbonyl group — perpendicular attack, tetrahedral alkoxide, protonation.

STEP 1 — attack R H C δ+ O δ− Nu STEP 2 — alkoxide C H R Nu O +H⁺ STEP 3 — product C H R Nu OH

Under basic conditions the free nucleophile attacks first and the alkoxide is protonated afterwards. Under acidic conditions, NIOS (§27.1.4) notes the order is reversed: the carbonyl oxygen is protonated first to give an oxocarbenium ion ($\ce{>C=\overset{+}{O}H}$), which makes the carbon even more electrophilic, and only then does the (often weaker) nucleophile add. Either way the carbon ends up tetrahedral with $\ce{Nu}$ and $\ce{OH}$ across the former double bond.

Reactivity order: aldehydes > ketones

Aldehydes are generally more reactive than ketones towards nucleophilic addition for two cooperating reasons that NCERT and NIOS both stress:

FactorAldehyde (one R or H)Ketone (two R)
Electronic (+I effect) Only one alkyl group donates electrons, so carbon stays strongly $\delta^+$ and electrophilic. Two alkyl groups feed electron density into the carbon, lowering its electrophilicity.
Steric (crowding) One small substituent leaves the carbon open to attack. Two bulky groups crowd the carbon and block the incoming nucleophile.
Net result Faster addition, equilibrium lies further right. Slower addition, equilibrium often lies left.

Putting numbers to the trend: formaldehyde $\ce{HCHO}$ has no electron-donating alkyl group and no steric bulk, so it is the most reactive carbonyl of all. The standard NEET reactivity order is therefore:

$$\ce{HCHO} > \ce{CH3CHO} > \ce{CH3COCH3} > \ce{CH3COCH2CH3}$$
Figure 2

How a second alkyl group dampens electrophilicity and crowds the carbonyl carbon.

HCHO C δ+++ H H O most reactive R–CHO C δ++ R → H O R–CO–R' C δ+ R → R' → O least reactive reactivity towards nucleophiles decreases
NEET Trap

Benzaldehyde is less reactive than propanal

Students assume an aldehyde always beats anything with extra carbons. But in benzaldehyde the ring donates electron density into the carbonyl by resonance, reducing the $\delta^+$ on carbon. So benzaldehyde is less reactive than propanal. Likewise an electron-withdrawing group raises reactivity: $p$-nitrobenzaldehyde > benzaldehyde, while $p$-tolualdehyde and acetophenone fall below.

Order (NCERT Q 8.4): Acetophenone < $p$-tolualdehyde < benzaldehyde < $p$-nitrobenzaldehyde.

HCN addition: cyanohydrins

Aldehydes and ketones add hydrogen cyanide to give cyanohydrins, in which a $\ce{-OH}$ and a $\ce{-CN}$ end up on the same carbon. Pure HCN reacts very slowly, so the reaction is base-catalysed: the base generates cyanide ion $\ce{CN^-}$, a much stronger nucleophile, which adds readily to the carbonyl carbon.

$$\ce{CH3COCH3 + HCN ->[\text{base}] (CH3)2C(OH)CN}$$

The product carries one carbon more than the starting carbonyl compound, which is why cyanohydrins are valued synthetic intermediates: their nitrile group can be hydrolysed to a $\ce{-COOH}$, giving an $\alpha$-hydroxy acid (NIOS §27.2.2). NEET frequently chains these two steps.

NEET Trap

When steric bulk kills the cyanohydrin

Cyclohexanone forms its cyanohydrin in good yield, but 2,2,6-trimethylcyclohexanone does not (NCERT Q 8.18). The three methyl groups flanking the carbonyl block the approach of $\ce{CN^-}$. The lesson: nucleophilic addition is exquisitely sensitive to crowding near the carbon.

Reactivity towards HCN (NCERT Q 8.12): di-tert-butyl ketone < methyl tert-butyl ketone < acetone < acetaldehyde.

Sodium bisulphite addition

Sodium hydrogensulphite ($\ce{NaHSO3}$) adds to aldehydes and ketones to give a crystalline bisulphite addition product. The position of equilibrium lies largely to the right for most aldehydes (and methyl ketones) but to the left for most ketones because of steric reasons.

$$\ce{>C=O + NaHSO3 <=> >C(OH)(SO3Na)}$$

The addition compound is water-soluble and can be filtered off, then regenerated to the original carbonyl compound by treatment with dilute mineral acid or alkali. This reversibility makes the bisulphite reaction a classic method to separate and purify aldehydes from non-carbonyl impurities.

Keep building the chapter

The same $\alpha$-acidity that drives the next reaction class powers the aldol condensation and α-hydrogen reactions — read that next.

Grignard reagents and water

Grignard reagents ($\ce{RMgX}$) are powerful carbanion-like nucleophiles. The $\ce{R^-}$ adds to the carbonyl carbon to give an alkoxide, which on acidic work-up ($\ce{H3O+}$) yields an alcohol. The carbonyl partner decides the class of alcohol formed (NIOS §27.1.4):

Carbonyl partnerAdds RMgX, then H₃O⁺ givesClass
Methanal, HCHORCH2OHPrimary alcohol
Any other aldehyde, R'CHOR'CH(OH)RSecondary alcohol
Ketone, R'COR''R'R''C(OH)RTertiary alcohol

Water itself is the simplest nucleophile here. Addition of water across $\ce{C=O}$ gives a gem-diol (hydrate), $\ce{>C(OH)2}$. For most carbonyls the equilibrium lies far to the left, but for very electrophilic carbonyls — formaldehyde and trichloroacetaldehyde — the hydrate is favoured. The same lone-pair-on-oxygen nucleophile reappears, more usefully, when the oxygen comes from an alcohol.

Alcohols: hemiacetals and acetals

An aldehyde reacts with one equivalent of a monohydric alcohol in the presence of dry HCl gas to give an hemiacetal — a carbon bearing one $\ce{-OH}$ and one $\ce{-OR}$. With a second molecule of alcohol the $\ce{-OH}$ is replaced by another $\ce{-OR}$, giving a gem-dialkoxy compound, the acetal.

$$\ce{CH3CHO ->[CH3OH][\text{dry } HCl] CH3CH(OH)(OCH3) ->[CH3OH][\text{dry } HCl] CH3CH(OCH3)2}$$

Ketones behave analogously to give ketals; with ethylene glycol they form cyclic ethylene glycol ketals. The dry HCl protonates the carbonyl oxygen, raising the electrophilicity of the carbon and so facilitating attack by the alcohol oxygen. Acetals and ketals are stable in base but are hydrolysed back to the parent carbonyl by aqueous mineral acid — which is why they are used as protecting groups for the carbonyl function.

NEET Trap

Hemiacetal vs acetal — count the OR groups

A hemiacetal has one $\ce{-OH}$ + one $\ce{-OR}$ on the same carbon (one molecule of alcohol added). An acetal has two $\ce{-OR}$ groups on that carbon (two alcohol molecules). "Hemi" means half — it is the half-way intermediate. Mixing these up is a frequent NEET error in structure-matching items.

Ammonia derivatives: addition-elimination

Nucleophiles of the type $\ce{H2N-Z}$ — ammonia and its derivatives — add to the carbonyl, but the initial addition product (a carbinolamine) is unstable and rapidly loses water. The equilibrium is therefore driven towards a stable product carrying a carbon–nitrogen double bond, $\ce{>C=N-Z}$. Because addition is followed by elimination of water, this class is called nucleophilic addition-elimination (a condensation). The reaction is reversible and acid-catalysed.

$$\ce{>C=O + H2N-Z <=>[\text{H+}] >C=N-Z + H2O}$$
Figure 3

Addition-elimination: the nucleophile adds, then water leaves to give the stable >C=N–Z.

C O + H₂N–Z add OH C NHZ unstable carbinolamine – H₂O C N Z stable >C=N–Z

The identity of $\ce{Z}$ names the product. NCERT Table 8.2 catalogues the family that NEET examines repeatedly:

Reagent (H₂N–Z)ZProduct (>C=N–Z)
Ammonia–HImine
Primary amine–RSubstituted imine (Schiff's base)
Hydroxylamine–OHOxime
Hydrazine–NH2Hydrazone
Phenylhydrazine–NHC6H5Phenylhydrazone
2,4-Dinitrophenylhydrazine–NHC6H3(NO2)22,4-DNP derivative (yellow–orange–red solid)
Semicarbazide–NHCONH2Semicarbazone

Two practical points NEET loves. First, the 2,4-DNP (Brady's) derivatives are crystalline solids with sharp melting points, used to characterise and identify aldehydes and ketones — a positive orange-red precipitate simply confirms a $\ce{>C=O}$. Second, although semicarbazide has two $\ce{-NH2}$ groups, only the one not attached to the carbonyl-like $\ce{C=O}$ of the urea fragment is nucleophilic enough to react, so only one participates in semicarbazone formation (NCERT Q 8.18).

Quick Recap

Nucleophilic addition in one screen

  • Carbonyl carbon is $\delta^+$ and electrophilic; nucleophile attacks perpendicular to the $sp^2$ plane → tetrahedral alkoxide → protonation → neutral product.
  • Reactivity $\ce{HCHO} > \ce{CH3CHO} > $ ketones, from less +I donation and less steric crowding in aldehydes.
  • HCN (base-catalysed) → cyanohydrin (one extra C, hydrolyses to α-hydroxy acid).
  • $\ce{NaHSO3}$ → water-soluble crystalline adduct; reversible, used to purify aldehydes.
  • Grignard then $\ce{H3O+}$ → 1°/2°/3° alcohol; water → gem-diol (hydrate).
  • Alcohol + dry HCl → hemiacetal (one OR) then acetal (two OR); acid-hydrolysable protecting group.
  • $\ce{H2N-Z}$ → addition-elimination → >C=N–Z: imine, Schiff's base, oxime, hydrazone, 2,4-DNP, semicarbazone.

NEET PYQ Snapshot — Nucleophilic Addition Reactions of Aldehydes & Ketones

Real NEET questions from the official papers, mapped to the addition reactions above.

NEET 2022 · Q.77

Match the products formed with the reagent (carbonyl reacts with): (a) Cyanohydrin (b) Acetal (c) Schiff's base (d) Oxime — against (i) NH₂OH, (ii) RNH₂, (iii) alcohol, (iv) HCN.

(1) a-ii, b-iii, c-iv, d-i  ·  (2) a-i, b-iii, c-ii, d-iv  ·  (3) a-iv, b-iii, c-ii, d-i  ·  (4) a-iii, b-iv, c-ii, d-i

Answer: (3)

Cyanohydrin ← HCN; acetal ← alcohol; Schiff's base ← RNH₂ (primary amine); oxime ← NH₂OH (hydroxylamine). This is a direct test of the addition and addition-elimination products tabulated above.

NEET 2023 · Q.67

A carbonyl compound reacts with HCN, then the product is treated with $\ce{OH^-}$ and finally with conc. $\ce{H2SO4}$. The final product [C] is:

(1) –COOH  ·  (2) –OH  ·  (3) –COOH  ·  (4) –CHO

Answer: (1) carboxylic acid

HCN adds to $\ce{C=O}$ to give the cyanohydrin (a $\ce{-CN}$ + $\ce{-OH}$ on one carbon); hydrolysis of the nitrile then yields the carboxylic acid. This is exactly the "cyanohydrins are synthetic intermediates" route — addition followed by nitrile hydrolysis.

NEET 2016 · Q.36

The product formed by the reaction of an aldehyde with a primary amine is:

(1) Ketone  ·  (2) Carboxylic acid  ·  (3) Aromatic acid  ·  (4) Schiff base

Answer: (4) Schiff base

A primary amine ($\ce{H2N-R}$) undergoes nucleophilic addition-elimination with the carbonyl to give a substituted imine, $\ce{>C=N-R}$, known as a Schiff's base — water is eliminated from the carbinolamine intermediate.

FAQs — Nucleophilic Addition Reactions of Aldehydes & Ketones

The conceptual sticking points NEET aspirants ask most about this topic.

Why do aldehydes and ketones undergo nucleophilic addition while alkenes undergo electrophilic addition?

In the carbonyl group the C=O bond is polarised because oxygen is far more electronegative than carbon, leaving the carbonyl carbon electron-deficient (delta-positive) and electrophilic. A nucleophile is therefore attracted to that carbon. In an alkene the C=C bond is non-polar and electron-rich, so it instead attracts electrophiles. The opposite polarity of the two double bonds is exactly why their addition chemistry runs in opposite directions.

Why are aldehydes more reactive than ketones in nucleophilic addition?

Two reasons, both favouring the aldehyde. Electronically, a ketone carries two alkyl groups whose +I (electron-releasing) effect feeds electron density into the carbonyl carbon and lowers its electrophilicity; an aldehyde has only one such group, so its carbon stays more positive. Sterically, the two bulky substituents in a ketone hinder the nucleophile's approach, whereas an aldehyde with one small substituent (often H) is far more accessible. Hence the order HCHO > CH3CHO > CH3COCH3.

Why is the sodium bisulphite addition test used for aldehydes but not most ketones?

Sodium hydrogensulphite adds across C=O to give a water-soluble crystalline bisulphite addition compound. The equilibrium lies well to the right for aldehydes and for most methyl ketones, but for bulkier ketones steric crowding pushes it to the left, so little product forms. Because the solid can be filtered off, washed, and then regenerated to the pure carbonyl compound by dilute acid or alkali, the test is a standard way to separate and purify aldehydes.

What is the difference between a hemiacetal and an acetal?

When an aldehyde adds one molecule of alcohol in dry HCl it gives a hemiacetal, which has one -OH and one -OR group on the same carbon. With a second molecule of alcohol the -OH is replaced by another -OR, giving an acetal, which bears two -OR groups on that carbon. Ketones similarly give ketals; with ethylene glycol they form cyclic ethylene glycol ketals. Acetals and ketals are stable to base but are hydrolysed back to the carbonyl compound by aqueous mineral acid, which is why they serve as protecting groups.

Why does the reaction of carbonyl compounds with ammonia derivatives stop at >C=N–Z and not at the addition product?

Ammonia derivatives H2N-Z first add to C=O to give an unstable carbinolamine (tetrahedral addition product carrying -OH and -NHZ). This intermediate rapidly loses water, and the resulting >C=N-Z is far more stable, so the equilibrium is driven towards the dehydrated product. The sequence is therefore classed as nucleophilic addition-elimination (condensation) rather than simple addition. Products include imines (Z=H, R), oximes (Z=OH), hydrazones (Z=NH2), 2,4-DNP derivatives, and semicarbazones (Z=NHCONH2).

Why is the 2,4-DNP (Brady's reagent) test useful in NEET problems?

2,4-Dinitrophenylhydrazine reacts with both aldehydes and ketones to give 2,4-dinitrophenylhydrazones, which separate as yellow, orange or red crystalline solids with sharp melting points. A positive orange-red precipitate confirms a carbonyl group (>C=O), so in identification problems it is the first clue that a compound is an aldehyde or ketone. It does not distinguish aldehydes from ketones — for that you use Tollens' or Fehling's test.